Responding to a pm:

You need to understand the concepts of mean, median and range very well to do this question. Since you asked me to resolve it, I am assuming that you are comfortable with these stats concepts I have discussed on my blog. Here is the logical solution:

"the difference between the highest and lowest salaries is $100,000."
So there are at least 2 people - say one with salary 0 and the other with 100k. No salary will be outside this range.

Median = 50k more than lowest. So median is right in the center of lowest and highest since lowest and highest differ by 100k. In our example, median = 50k. Since there are more than 2 people, there would probably be a person at 50k.

Mean = 20k more than median so in our example, mean salary = 70k

On the number line,

0........50k (median)........100k

Mean = 70k

So there must be people more toward 100k to bring the mean up to 70k. Since we want to add minimum people, we will add people at 100k to quickly make up the right side deficit. 0 and 50k are (70k + 20k) = 90k away from 70k. 100k is 30k away from 70k. To bring the mean to 70k, we will add two people at 100k each to get:

0....50k.....100k, 100k, 100k

But when we add more people to the right of 70k, the median will shift to the right. We need to keep the median at 50k. So every time we add people to the right of 70k, we need to add people at 50k too to balance the median. 50k is 20k less than 70k while 100k is 30k more than 70k. To keep the mean same, we need to add 2 people at 100k for every 3 people we add at 50k. So if we add 3 people at 50k and 2 people at 100k, we get:

0, ... 50k, 50k, 50k, 50k, ... 100k, 100k, 100k, 100k, 100k
the median is not at 50k yet.

Add another 3 people at 50k and another 2 at 100k to get

0, 50k, 50k, 50k, 50k, 50k, 50k, 50k, 100k, 100k, 100k, 100k, 100k, 100k, 100k

Now the median is 50k and mean is 70k.

Total number of people is 15.

Answer (C)
_________________

As for your question as to how close this question is to actual GMAT questions, quite close I would say. You need a solid understanding of very basic concepts and need to just apply those. The question is easily do-able within 2 mins. I can't find any reason to call this is a 'non-GMAT like' question.
_________________

Hello Karishma,

That is a great solution. Unfortunately, I am unable to understand the concept behind this. Kindly give the link of your blog as reference. That would be beneficial for me.

Here are some posts discussing mean and median:

http://www.veritasprep.com/blog/2012/04 ... etic-mean/
http://www.veritasprep.com/blog/2012/04 ... questions/
http://www.veritasprep.com/blog/2012/05 ... eviations/
http://www.veritasprep.com/blog/2012/05 ... tic-means/
http://www.veritasprep.com/blog/2012/05 ... on-median/
_________________

We can also use POE .Median is the middle number between lowest and highest so total number of students will be odd .This eliminates choices A,B,D
Between C and E .Average is 70000 which is more towards higher pay and is also more then median .So number of students will be more on the rhs then lhs .Add the students accordingly on both sides to get the final answer.

Regards,
Manish Khare
_________________

This question is definitely a good question on statistics. Essentially, it tests your in depth knowledge of averages in general and weighted average in particular.

A smart approach in this question, as highlighted by some experts already on this thread, is to take values that will make problem solving simpler.

The mean salary is higher than the median salary, which points out to the fact that, the number of values greater than the mean and closer to the highest salary, is higher.

Let’s take the lowest salary be $0. If we take 10 students, the tenth value should be $100000. The median, which is the average of the fifth and sixth value should be $50000. So, the list can look like this:
0, 50, 50, 50, 50, 50, 100,100,100,100.

The idea is to take every value to the left of the median as $50000 to ensure the median is $50000 and to take every value to the right of the median as $100000. However, these values do not total up to $700000 [ $70000 (mean) * 10 (number of employees)]. Notice that the number of 50s are higher than the 100s and hence the mean is not 70.

A similar thing happens when we take 12 values:
0, 50, 50, 50 , 50, 50, 50, 100,100,100, 100, 100. The mean for this set comes out to be 66.66.

When we take 15 values, the median is the 8th value. So, we can have a 0, seven 50s and seven 100s. This gives us a mean of 70 and also ensures that the median is 50 along with the range being 100.

The correct answer option is C.

Hope this helps!
_________________

Let the lowest salary be x. Then, the median salary is x + 50,000; the average salary is x + 70,000 and the greatest salary is x + 100,000.

We notice that the average salary is between the median and the greatest salary. When this is the case, the number of students in the class can be minimized by choosing the salaries as one salary which is equal to x, some number of salaries which are equal to x + 50,000 and some number of salaries which are equal to x + 100,000. The reason for such a choice of salaries is because every salary we add that is less than x + 50,000 will bring the average salary down, and to counter that, since we can’t add any salaries greater than x + 100,000, we have to increase the number of salaries that equal x + 100,000, therefore increasing the number of students.

So, let n be the number of salaries that equal x + 50,000 and m be the number of salaries that equal x + 100,000. Notice that m cannot be greater than n because otherwise, the median will equal x + 100,000. Since the average salary is x + 70,000 and the number of students is n + m + 1, the sum of all the salaries is (n + m + 1)(x + 70,000). On the other hand, the sum of all the salaries is also equal to x + n(x + 50,000) + m(x + 100,000). Let’s equate these two quantities:

x + n(x + 50,000) + m(x + 100,000) = (n + m + 1)(x + 70,000)

x + nx + 50,000n + mx + 100,000m = nx + mx + x + 70,000n + 70,000m + 70,000

30,000m = 20,000n + 70,000

3m = 2n + 7

Since m ≤ n, 3m ≤ 3n. Let’s substitute 3m = 2n + 7:

2n + 7 ≤ 3n

7 ≤ n

We see that the smallest value of n is 7. If n = 7, then 3m = 14 + 7 = 21; therefore m = 7 as well. Thus, the smallest number of students is n + m + 1 = 7 + 7 + 1 = 15.

Answer: C
_________________

I did not understand how 0k and 50k are 90k away from 70k, could you please help in clearing this doubt?

Soln

Let’s take the lowest salary to be $0. The Median salary is $ 50,000 & the Highest salary is $100000.

3 element- 0, 50000 , 100,000 but in this case Avg is 150,000/3 = 50,000

To increase the Avg up to 70000 keeping median at 50,000, we have to add max value to left & right of the median

By adding 1-1 element both side of the median

0, 50k, 50 k, 100k,100 k (As we have to take minimum no. of the student to reach the 70 k Avg we have to add maximum possible value to the left & right of the median).

note that the Avg of 5 student's salary is - 300k/5= 60 k , to increase the avg to 70 k we have to add more student with salary $50 k & $100 k to both left & right of the median.

As "0" is the salary of 1st & only student: The no of the student with salary $0 is always 1

according to our observation, We can infer that there would be "x" no of the student with salary 50 k & "x" no. of student with salary 100 k, to get avg salary of $70 k.

Total student - 2x+1

Total salary- 0+50x+100x= 150x...... (1)

Desired total - total student * with avg salary 70 k = (2x+1)*70.... (2)

equating total salary with the desired total ( by eq 1 &2)
150x=(2x+1)*70 ===> x= 7 where x is no is student with salary 50 k & same no. of student with salary 100 k

total no. of student = 2x+1 = 2*7+1 = 15

Option D is correct answer

Hello VeritasKarishma, could the problem also be solved as I've outlined below?
First off, we've got our initial 3 salaries -> 0...50k...100k
Now, the average is 70; the elements below the average constitute a deficit of 90k and the elements above a surplus of 30k.
Considering weighted averages, w1/w2 = 30/90 = 3/9. Can one conclude from this that 12 elements need to be added to the existing 3, leading to a total of 15 students? If not, could you please explain why?

You have not taken the median into consideration. The median needs to be at 50k. That is the reason we need to add 12 more people. Otherwise, w1/w2 = 3/9 = 1/3. We already have 1 person at 100k so adding 2 more people at 100k should have been enough.
Btw, good you are thinking in these terms.
_________________

Lets assume the number of students be n


1st case n = odd


then ( n + 1 )/2 th number is 50k + L , in the set [ L , H ] where H and L are highest and lowest salary.

therefore the set becomes

L ---- ( n-1 )/2 terms ---- [ (the Median = 50k + L ) (n + 1)/2 th term ----- ( n-1 )/2 terms ------ H = L+100k

now only two of the options 15 and 25 mentioned in the question satisfies ( n -1 )/2 because they are odd and the minimum of them is 15.


1st case n = odd


then ( n + 1 )/2 th number is 50k + L , in the set [ L , H ] where H and L are highest and lowest salary.

therefore the set becomes

L ---- ( n-1 )/2 terms ---- [ (the Median = 50k + L ) (n + 1)/2 th term ----- ( n-1 )/2 terms ------ H = L+100k

now only two of the options C = 15 and D = 25 mentioned in the question satisfies ( n -1 )/2 because they are odd and the minimum of them is 15.


2nd case n = Even

then Avg of ( n )/2th term and (n/2 +1)th term is the Median = 50k + L , in the set [ L , H ] where H and L are highest and lowest salary.

therefore the set becomes

L ---- ( n-2 )/2 terms ---- [ (the Median = 50k + L ) ( 2 terms are taken) term ----- ( n -2 )/2 terms ------ H = L+100k

now three options A, B, D mentioned in the question satisfies ( n - 2 )/2 because they are even and the minimum of them is A.

The best method to solve this in 2 min time frame is ...take median and mean simultaneously, since we need to keep the median as 50 and average as 70.
We need to have series in which the sum must be multiple of 350.
ex. 0,50,50,50,100,100
now because we need to have sum a multiple of 70 and median as 50. we need to find n*350 for which this is satisfied.
we can check for 350,700,1050

for 1050 all statements are satisfied.